Continuous-time Fourier Methods

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1 ELEC SIGNALS and SYSTEMS Continuous-time Fourier Methods Chapter 6 1

2 Representing a Signal The convolution method for finding the response of a system to an excitation takes advantage of the linearity and timeinvariance of the system and represents the excitation as a linear combination of impulses and the response as a linear combination of impulse responses The Fourier series represents a signal as a linear combination of complex sinusoids 2

3 Linearity and Superposition If an excitation can be expressed as a sum of complex sinusoids the response of an LTI system can be expressed as the sum of responses to complex sinusoids. 3

4 Real and Complex Sinusoids cos( x) = e jx + e - jx 2 sin( x) = e jx - e - jx j2 4

Jean Baptiste Joseph Fourier Joseph Fourier was a French mathematician and physicist born in Auxerre and best known for initiating the investigation of Fourier series and their applications to

5 Jean Baptiste Joseph Fourier Joseph Fourier was a French mathematician and physicist born in Auxerre and best known for initiating the investigation of Fourier series and their applications to problems of heat transfer and vibrations. The Fourier Transform and Fourier's law are also named in his honour. Fourier is also generally credited with the discovery of the greenhouse effect. 3/21/1768-5/16/1830 5

6 Conceptual Overview 6

7 Conceptual Overview The Fourier series represents a signal as a sum of sinusoids. The best approximation to the dashed-line signal below using only a constant is the solid line. (A constant is a Constant cosine of zero frequency.) t -0.6 x(t) 1.6 Exact x(t) Approximation of x(t) by a constant t 0 t 0 + T t 7

8 Conceptual Overview The best approximation to the dashed-line signal using a constant plus one real sinusoid of the same fundamental frequency as the dashed-line signal is the solid line. Sinusoid t -0.6 x(t) 1.6 Exact x(t) Approximation of x(t) through 1 sinusoid t 0 t 0 + T t 8

9 Conceptual Overview The best approximation to the dashed-line signal using a constant plus one sinusoid of the same fundamental frequency as the dashed-line signal plus another sinusoid of twice the fundamental frequency of the dashed-line signal is the solid line. 9

10 Conceptual Overview The best approximation to the dashed-line signal using a constant plus three sinusoids is the solid line. In this case (but not in general), the third sinusoid has zero amplitude. This means that no sinusoid of three times the fundamental frequency improves the approximation. Sinusoid t -0.6 x(t) 1 Exact x(t) Approximation of x(t) through 3 sinusoids t 0 t 0 + T t 10

11 Conceptual Overview The best approximation to the dashed-line signal using a constant plus four sinusoids is the solid line. This is a good approximation which gets better with the addition of more sinusoids at higher integer multiples of the fundamental frequency. 11

12 Continuous-Time Fourier Series Definition The Fourier series representation of a signal x(t) over a time t 0 < t < t 0 + T is ( ) = c x k x t å k=- [ ]e j 2p kt /T where c x [k] is the harmonic function and k is the harmonic number. The harmonic function can be found from the signal using the princple of orthogonality. 12

13 Orthogonality 13

14 Orthogonality Using Euler's identity t 0 +T ( e j 2p kt /T j 2pqt /T,e ) é æ = ò cos 2p k - q ê è ç ë T If k = q, t 0 t 0 +T ( ) + j sin( 0) t ö æ ø + j sin 2p k - q è ç T t ö ù ø ú û dt ( e j 2p kt /T j 2pqt /T,e ) = ò éë cos 0 ùû dt = ò dt = T. If k ¹ q, the integral ( e j 2p kt /T j 2pqt /T,e ) é æ = ò cos 2p k - q ê è ç ë T t 0 t 0 +T t 0 t 0 +T t 0 t ö æ ø + j sin 2p k - q è ç T t ö ù ø ú û dt is over a non-zero integer number of cycles of a cosine and a sine and is therefore zero. 14

15 Orthogonality Therefore e j 2p kt /T and e j 2pqt /T are orthogonal if k and q are not equal. Now multiply the Fourier series expression x t - j 2pqt /T by e (q an integer) å k=- x( t)e - j 2pqt /T = c x [ k]e j 2p ( k-q)t /T å k=- ( ) = c x k and integrate both sides over the interval t 0 t < t 0 + T t 0 +T é ò x( t)e - j 2pqt /T j 2p ( k-q)t /T ù dt = ò ê å c x [ k]e ú ëk=- û dt. t 0 t 0 +T t 0 [ ]e j 2p kt /T 15

16 Orthogonality 16

17 Continuous-Time Fourier Series Definition 17

18 CTFS of a Real Function It can be shown that the continuous-time Fourier series (CTFS) harmonic function of any real-valued function x t that c x [ k] = c * x [-k]. ( ) has the property One implication of this fact is that, for real-valued functions, the magnitude of the harmonic function is an even function and the phase is an odd function. 18

19 The Trigonometric CTFS The fact that, for a real-valued function x t ( ) c x [ k] = c * x [-k] also leads to the definition of an alternate form of the CTFS, the so-called trigonometric form. x( t) = a x 0 where [ ] + a x k a x å k=1 [ ]cos 2pkt / T ( ) + b x k [ ]sin 2pkt / T ( ) { } [ k] = 2 T t 0 +T ò t 0 x( t)cos( 2pkt / T )dt b x [ k] = 2 T t 0 +T ò t 0 x( t)sin( 2pkt / T )dt 19

20 The Trigonometric CTFS Since both the complex and trigonometric forms of the CTFS represent a signal, there must be relationships between the harmonic functions. Those relationships are ì ï ï í ï ï î b x a x [ 0] = c x [ 0] b x [ 0] = 0 [ k] = c x [ k] + c * x [ k] [ k] = j c x [ k] - c * x [ k] a x ì ï ï í ï ï c x -k îï c x ( ) ü ï ï ý ï ï þ c x [ 0] = a x [ 0] [ k] = a k x [ ] - j b x [ k] [ ] = c * x [ k] = a k x [ ] + j b x [ k] 2 2, k = 1,2,3, ü ï ï ý ï ï þï, k = 1,2,3, 20

21 CTFS Example #1 21

22 CTFS Example #1 22

23 CTFS Example #2 Let a signal be defined by x t ( ) = 2cos 400pt ( ) and let T = 10 ms which is 2T 0. 23

24 CTFS Example #2 24

25 CTFS Example #3 Let x t ( ) = 1/ 2-3 / 4 ( )cos 20pt ( ) + 1/ 2 ( )sin 30pt ( ) and let T = 200 ms. 25

26 CTFS Example #3 26

27 CTFS Example #3 27

28 CTFS Example #3 28

29 Linearity of the CTFS These relations hold only if the harmonic functions of all the component functions are based on the same representation time T. 29

30 CTFS Example #4 Let the signal be a 50% duty-cycle square wave with an amplitude of one and a fundamental period T 0 = 1. x( t) = rect ( 2t)*d 1 ( t) 30

31 CTFS Example #4 31

32 CTFS Example #4 32

33 CTFS Example #4 33

34 CTFS Example #4 A graph of the magnitude and phase of the harmonic function as a function of harmonic number is a good way of illustrating it. 34

35 The Sinc Function 35

36 CTFS Example #5 Let x t ( ) = 2cos 400pt ( ) and let T = 7.5 ms which is 1.5 fundamental periods of this signal. 36

37 CTFS Example #5 37

38 CTFS Example #5 38

39 CTFS Example #5 The CTFS representation of this cosine is the signal below, which is an odd function, and the discontinuities make the representation have significant higher harmonic content. This is a very inelegant representation. 39

40 CTFS of Even and Odd Functions For an x x k k k even function, the complex CTFS harmonic function x k c is purely real and the sine harmonic function b is zero. For an odd function, the complex CTFS harmonic function c is purely imaginary and the cosine harmonic function a x is zero. 40

41 Convergence of the CTFS For continuous signals, convergence is exact at every point. Partial CTFS Sums x N N å ( t) = c x [ k]e j 2pkt /T 0 k=- N A Continuous Signal 41

42 Convergence of the CTFS Partial CTFS Sums For discontinuous signals, convergence is exact at every point of continuity. Discontinuous Signal 42

43 Convergence of the CTFS At points of discontinuity the Fourier series representation converges to the mid-point of the discontinuity. There is an obvious overshoot and ripple near the discontinuities that does not appear to become smaller as N increases. In fact, the maximum vertical overshoot near a discontinuity does not decrease with N, even as N approaches infinity. This overshoot is called the Gibbs phenomenon 43

44 Numerical Computation of the CTFS How could we find the CTFS of a signal which has no known functional description? Numerically. c x [ k] = 1 T ò T x( t)e - j 2pkt /T dt Unknown 44

45 Numerical Computation of the CTFS c x [ 1 T N -1 å n=0 é ê ëê ( n+1)t s ò nt s ù x( nt s )e - j2p knt s /T dt ú ûú Samples from x(t) 45

46 Numerical Computation of the CTFS It can be shown that, for harmonic numbers k where N x DFT c k 1/ N x nt, k N DFT N -1 x x nts nts e n 0 s - j2 nk / N 46

47 Numerical Computation of the CTFS 47

48 CTFS Properties Let a signal x(t) have a fundamental period T 0 x and let a signal y(t) have a fundamental period T 0 y. Let the CTFS harmonic functions, each using a common period T as the representation time, be c x [k] and c y [k]. Then the following properties apply. Linearity 48

49 CTFS Properties Time Shifting 49

50 CTFS Properties Frequency Shifting (Harmonic Number Shifting) A shift in frequency (harmonic number) corresponds to multiplication of the time function by a complex exponential. Time Reversal 50

51 CTFS Properties Time Scaling Let z t ( ) = x( at), a > 0 ( ) [ k] = c x [ k] ( ) Case 1. T = T 0 x / a = T 0z for z t c z Case 2. T = T 0 x for z t If a is an integer, c z [ ], k / a an integer ì [ k] = c x k / a í î0, otherwise 51

52 CTFS Properties Time Scaling (continued) 52

53 CTFS Properties 53

54 CTFS Properties Change of Representation Time 54

55 CTFS Properties Time Differentiation 55

56 CTFS Properties Time Integration Case 1. c x [ 0] = 0 Case 1 Case 2 Case 2. c x [ 0] ¹ 0 t ò - x( l)dl is not periodic 56

57 CTFS Properties 57

58 CTFS Properties 58

59 CTFS Properties 59

60 Some Common CTFS Pairs 60

61 CTFS Examples Find the CTFS harmonic function of x( t) with T = c x c x c x c x c x c x [ k] = ( 1 / T ) x t 10-8 ò ( )e - j 2p kt /T dt Þ c x 0 T 10-8 ò ( )dt = 35 / 2 [ ] = t [ k] = 10 8 ò ( t)e - j 2p 108 kt dt = ò te - j 2p 108 kt dt ì é [ k] = ï íêt îï ë 0 e - j 2p 108 kt - j2p 10 8 k ù ú û ì [ k] = e- j 2p k 10-8 j2p 10 8 k - e - j 2p 10 8 kt ï é í ê ï ê ë j2p 10 8 k î [ k] = e - j 2p k é ê + ë j2pk ì1 / 2, k = 0 ï [ k] = 35 í j î ï2pk, k ¹ 0 ò 0 æ ç è ( ) 2 0 e - j 2p 108 kt - j2p 10 8 k ù ú ú û ü ï ý ï þ ö ü ø dt ï ý þï 1- e - j 2p k ù ( j2pk) û ú = e- j 2p k - j2pke - j 2p k ( j2pk) 2 61

62 Continuous-Time Fourier Transform 62

63 Extending the CTFS The CTFS is a good analysis tool for systems with periodic excitation but the CTFS cannot represent an aperiodic signal for all time The continuous-time Fourier transform (CTFT) can represent an aperiodic signal for all time 63

64 CTFS-to-CTFT Transition Consider a periodic pulse-train signal x( t) with duty cycle w / T 0 Its CTFS harmonic function is c x [ k] = Aw T 0 æ sinc kw ö è ç ø As the period T 0 is increased, holding w constant, the duty cycle is decreased. When the period becomes infinite (and the duty cycle becomes zero) x t T 0 ( ) is no longer periodic. 64

65 CTFS-to-CTFT Transition [ k] for 50% and 10% duty Below are graphs of the magnitude of c x cycles. As the period increases the sinc function widens and its magnitude falls. As the period approaches infinity, the CTFS harmonic function becomes an infinitely-wide sinc function with zero amplitude. w = T 0 2 w = T

66 CTFS-to-CTFT Transition This infinity-and-zero problem can be solved by normalizing the CTFS harmonic function. Define a new modified CTFS harmonic function T 0 c x [ k] = Awsinc( wkf 0 ) and graph it ( ) versus kf 0 instead of versus k. f 0 = 1/ T 0 66

67 CTFS-to-CTFT Transition In the limit as the period approaches infinity, the modified CTFS harmonic function approaches a function of continuous frequency f (kf 0 ). 67

68 CTFS-to-CTFT Transition 68

69 Definition of the CTFT 69

70 Some Remarkable Implications of the Fourier Transform The CTFT expresses a finite-amplitude, real-valued, aperiodic signal which can also, in general, be time-limited, as a summation (an integral) of an infinite continuum of weighted, infinitesimalamplitude, complex sinusoids, each of which is unlimited in time. (Time limited means having non-zero values only for a finite time. ) 70

71 Frequency Content Lowpass Highpass Bandpass 71

72 Some CTFT Pairs 72

73 Negative Frequency This signal is obviously a sinusoid. How is it described mathematically? It could be described by x t But it could also be described by x t ( ) = Acos( 2p f 0 t) ( ) = Acos 2pt / T 0 ( ) = Acos 2p - f 0 ( )t ( ) 73

74 Negative Frequency x(t) could also be described by or x( t) = A e j 2p f 0 t + e - j 2p f 0 t 2 x( t) = A 1 cos 2p f 0 t ( ) + A 2 cos 2p - f 0 ( )t ( ), A 1 + A 2 = A and probably in a few other different-looking ways. So who is to say whether the frequency is positive or negative? For the purposes of signal analysis, it does not matter. 74

75 Negative Frequency Consider an experiment in which we multiply two sinusoidal ( ) = cos 2p f 1 t ( ) = cos( 200pt) to form ( ) = x 1 t ( ) can be expressed using a trigonometric signals x 1 t x t ( )x 2 ( t). x t ( ) and x 2 t identity as x( t) = ( 1/ 2) é ë cos 2p ( f 1-100)t ( )t Now imagine that we continuously change ( ) + cos( 2p f ) f 1 from a frequency above100 to a frequency below 100. f becomes negative. ù û 75

76 Example 76

77 More CTFT Pairs The generalization of the CTFT allows us to extend the table of CTFT pairs to some very useful functions. 77

78 CTFT Properties Linearity 78

79 CTFT Properties 79

80 CTFT Properties x x Frequency Shifting j2 f0t F t e X f f0 j 0t F t e X 0 80

81 The Uncertainty Principle The time and frequency scaling properties indicate that if a signal is expanded in one domain it is compressed in the other domain. This is called the uncertainty principle of Fourier analysis. 81

82 CTFT Properties 82

83 CTFT Properties 83

84 CTFT Properties In the frequency domain, the cascade connection multiplies the frequency responses instead of convolving the impulse responses. 84

85 CTFT Properties 85

86 CTFT Properties 86

87 CTFT Properties 87

88 CTFT Properties 88

89 CTFT Properties X 0 ò - ( ) = x t ( )dt ( ) = X f x 0 ò - ( )df 89

90 CTFT Properties 90

91 Example Find CTFT of x(t)= 25 rect((t-4)/10) We can find the CTFT of the unit rectangle function in the table of Fourier Transforms rect(t) Linearity 25 rect(t) Time Scaling 25 rect(t/10) F F F sinc(f) 25 sinc(f) 250 sinc(10f) Time Shifting 25 rect((t-4)/10) F j 8 f 250 sinc(10f) e 91

Representing a Signal

The Fourier Series Representing a Signal The convolution method for finding the response of a system to an excitation takes advantage of the linearity and timeinvariance of the system and represents the